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This may be more of a math question, but I am trying to prove that for a function $f(q,\dot{q},t)$

$$\frac{d}{dt}\frac{∂f}{∂\dot{q}}=\frac{∂}{∂\dot{q}}\frac{df}{dt}−\frac{∂f}{∂q}.\tag{1}$$

As part of this problem, I have already proved that

$$\frac{d}{dt}\frac{\partial{f}}{\partial{q}}=\frac{\partial{}}{\partial{q}}\frac{df}{dt}\tag{2}$$

through the use of a total derivative expansion for $\frac{df}{dt}$, however I can't seem to figure out where the additional $-\frac{\partial{f}}{\partial{q}}$ term comes from for equation (1).

Qmechanic
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Synchronicity
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2 Answers2

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Careful with interchanging total and partial derivatives. In this context, I'm guessing you consider $f$ along a curve $q(t)$, then $$ \dfrac{d}{dt}f(q(t), \dot{q}(t), t) = \dfrac{\partial f}{\partial q}\dot{q}+ \dfrac{\partial f}{\partial \dot{q}}\ddot{q}+\dfrac{\partial f}{\partial t} $$ Something similar holds for $\tfrac{\partial f}{\partial \dot{q}}(q(t),\dot{q}(t), t)$. If you write that down explicitly and further compute $\tfrac{\partial}{\partial\dot{q}}\tfrac{df}{dt}$ you should see the result.

good luck.

KilianM
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OP's sought-for commutators $$\left[\frac{\partial}{\partial \dot{q}^j},\frac{\mathrm d}{\mathrm d t}\right]~\stackrel{(3)}{=}~\frac{\partial}{\partial q^j}\tag{1}$$ and $$\left[\frac{\partial}{\partial q^j},\frac{\mathrm d}{\mathrm d t}\right]~\stackrel{(3)}{=}~0\tag{2}$$ follow from the explicit formula $$\frac{\mathrm d}{\mathrm d t} ~=~\frac{\partial}{\partial t} +\dot{q}^j\frac{\partial}{\partial q^j} +\ddot{q}^j\frac{\partial}{\partial \dot{q}^j} +\dddot{q}^j\frac{\partial}{\partial \ddot{q}^j} +\ldots \tag{3}$$ for the total time derivative. See also e.g. this related Math.SE post & this related Phys.SE post.

Qmechanic
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